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Solvable GroupsLet us now introduce the class of groups known as solvable groups, which play a crucial role in the theory of the solution of equations in radicals. In some sense, one could say that the solvable groups are one order of magnitude more complicated than abelian groups. Before introducing the concept of a solvable group, we must define several preliminary notions. Throughout this section let G be a group of order n. Definition 1: A normal series for G is a chain of subgroups of G of the form
G = G_{0} G_{1} G_{2} ... G_{r} = {1}, G_{i} G_{i+i},
If two normal series for G are given
G = G_{0} G_{1} ... G_{r} = {1},
G = H_{0} H_{1} ... H_{s} = {1},
then the second is said to be a refinement of the first if every H_{i} is a G_{j} for some j. The second is said to be a proper refinement of the first if it is a refinement and Example 1: Let Example 2: Let
S_{3} A_{3} {1}.
We will show below that the normal series in Examples 1 and 2 are composition series. Definition 2: Let
G = G_{0} G_{1} G_{2} ... G_{r} = {1}
be a normal series for G. The groups (1)
G_{0}/G_{1}, G_{1}/G_{2} ... G_{r1}/G_{r}
are called composition factors of the normal series. If the series is a composition series, then the composition factors (1) are said to be composition factors for G. An important result about composition factors, which we will state but not prove is the theorem of Jordan and Hölder: Theorem 3: (JordanHölder): Let
G = G_{0} G_{1} ... G_{r} = {1},
G = H_{0} H_{1} ... H_{s} = {1},
be two composition series for G. Then
G_{j} = G_{j1}/G_{j} (j = 1,...,r)
can be rearranged, so that the ith composition factor is isomorphic to
H_{i} = H_{i1}/H_{i} (i = 1,...,r).
Thus, the composition factors of G are uniquely determined up to isomorphism and rearrangement, independent of the choice of the composition series. Recall that a simple group is a group with no nontrivial normal subgroups. Then the following result is very useful: Proposition 4: Let (2)
G = G_{0} G_{1} ... G_{r} = {1}
be a normal series for G This normal series is a composition series if and only if all composition factors are simple groups. Proof: Note that (2) is a composition series if and only if for every i Let us consider the normal series of Examples 1 and 2. Since and since Z_{2} and Z_{3} are simple groups, we see that the normal series of Examples 1 and 2 are composition series. Thus the composition factors for V_{4} are Z_{2} and Z_{2}, and the composition factors for S_{3} are Z_{2} and Z_{3}. Definition 5: G is said to be solvable if it has a composition series whose composition factors are cyclic of prime order. Proposition 6: G is solvable if and only if the composition factors of G are abelian. Proof: Obvious.
Let C be a composition factor of G. Then C is a simple, abelian group. Let p be a prime dividing C. Then C contains an element g of order p. Since C is simple and abelian, Example 3: V_{4} and S_{3} are solvable. Example 4: Let G be abelian. Then the composition factors of G are all abelian, so that G is solvable. Example 5: Let
S_{n} A_{n} {1}
are
S_{n}/A_{n} Z_{2}, A_{n}/{1} A_{n}.
Now A_{n} is simple by Abel's theorem, so that (3) is a composition series for S_{n}. However, for Example 6: Every group of prime power order is solvable Proposition 7: Let G be a solvable group, H a subgroup of G. Then H is solvable. Proof: Let
H_{i} = G_{i} H (0 < i < r).
Then
: G_{i1} H G_{i1}/G_{i},
(g) = gG_{i} (g G_{i1})
is a homomorphism with kernel
Proposition 8: Let G be a solvable group, H any group, Proof: Let
f(G) = f(G_{0}) f(G_{1}) ... f(G_{r}) {1}.
Moreover, for i = 1,...,r,
Corollary 9: Let G be a solvable group, Proof: Let f : G G/H be the natural homomorphism. Then 
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